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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>For <span class="process-math">\(r_1=i\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\left(
\begin{array}{cc}
2-i &amp; -5\\
1 &amp; -2-i
\end{array}
\right)
\left(
\begin{array}{c}
\xi^{(1)}_1\\
\xi^{(1)}_2
\end{array}
\right)={\bf 0},
\end{equation*}
</div>
<p class="continuation">which gives</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation}
\begin{array}{c}
(2-i) \xi^{(1)}_1-5 \xi^{(1)}_2=0\\
\xi^{(1)}_1-(2+i)  \xi^{(1)}_2=0.
\end{array}\tag{6.6.7}
\end{equation}
</div>
<p class="continuation">Note <span class="process-math">\((\ref{eq8_9})_1 \times (2+i)\)</span> leads to <span class="process-math">\(\xi^{(1)}_1-(2+i)  \xi^{(1)}_2=0\)</span> which is exactly <span class="process-math">\((\ref{eq8_9})_2\text{.}\)</span> From <span class="process-math">\((\ref{eq8_9})_2\text{,}\)</span> we let <span class="process-math">\(\xi^{(1)}_1=5\text{,}\)</span> then <span class="process-math">\(\xi^{(1)}_2=2-i\text{.}\)</span> So</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\vec{\xi}^{(1)}=\left(
\begin{array}{c}
5\\
2-i
\end{array}
\right)=\left(
\begin{array}{c}
5\\
2
\end{array}
\right)+i \left(
\begin{array}{c}
0\\
-1
\end{array}
\right)={\bf a}+i {\bf b}.
\end{equation*}
</div>
<span class="incontext"><a href="sec6_6.html#p-289" class="internal">in-context</a></span>
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